If $T$ is a symmetric bilinear form on vector space $V$, and let $U$ be a finite dimensional subspace of $V$, then $V=U+U^{\bot}$

Question

Here is the full question:

If $T$ is a symmetric bilinear form on vector space $V$, and let $U$ be a finite dimensional subspace of $V$, then $V=U+U^{\bot}$,

where $U^{\bot}$ is the orthogonal complement for $U$, given by $U^{\bot}=\{v \in V \mid T(u,v)=0\}$.

How does one prove that $V=U+U^{\bot}$?

We haven't introduced inner product spaces in class yet, and I understand that if $T$ was the inner product, standard proofs use: assume if $V \neq U + U^{\bot}$, there exists a vector $\alpha$ such it is orthogonal for all $\beta \in \text{span}(U,U^{\bot})$, implies $<\alpha, \beta>=0$, which implies $\alpha \in U^{\bot}$, contradiction.

But in this case, I cannot assume that $T$ is not an inner product? Are there any hints on how I can prove this statement?

Answer 1

I believe this is not true in general. Take $A=\left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right)$ and define $T(u,v)=u^T A v$, the symmetric bilinear form.

Let $W=\text{span}\{(1,-1,0)\}$. Then the only vectors that fulfils $(a,b,c)\left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right) \left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right)$

is such that $(a,b,c) \left(\begin{array}{r} 0 \\ 0 \\ 1 \end{array}\right)=0$ has dimension 2, but since every vector in $W$ satisfies this equation as well, $W \subset W^{\bot}$ and their sum cannot be possibly equal to $V$.